Energy released in fusion of `1 kg` of deuterium nuclei.

Options

A. `8 xx 10^13 J`
B. `6 xx 10^27 J`
C. `2 xx 10^7 KwH`
D. `8 xx 10^23 MeV`

Explanation

To calculate the energy released in the fusion of 1 kg of deuterium nuclei, we can follow these steps:

### Step 1: Understand the Fusion Reaction
The fusion of deuterium nuclei can be represented as:
\[ 2 \, ^2H \rightarrow \, ^4He + n + \text{Energy} \]
where \( ^2H \) is deuterium, \( ^4He \) is helium, and \( n \) is a neutron. The energy released in this reaction is approximately \( 3.27 \, \text{MeV} \) per reaction.

### Step 2: Convert Energy from MeV to Joules
1 MeV (mega electron volt) is equivalent to \( 1.6 \times 10^{-13} \) Joules. Therefore, the energy released in joules for one fusion reaction is:
\[ E = 3.27 \, \text{MeV} \times 1.6 \times 10^{-13} \, \text{J/MeV} = 5.232 \times 10^{-13} \, \text{J} \]

### Step 3: Calculate the Number of Deuterium Nuclei in 1 kg
To find the number of deuterium nuclei in 1 kg, we first calculate the number of moles of deuterium:
- The molar mass of deuterium (\( ^2H \)) is approximately \( 2 \, \text{g/mol} \).
- Therefore, the number of moles in 1 kg (1000 g) of deuterium is:
\[ \text{Number of moles} = \frac{1000 \, \text{g}}{2 \, \text{g/mol}} = 500 \, \text{mol} \]

### Step 4: Calculate the Total Number of Deuterium Nuclei
Using Avogadro's number (\( 6.022 \times 10^{23} \, \text{atoms/mol} \)), we can find the total number of deuterium nuclei:
\[ \text{Number of nuclei} = 500 \, \text{mol} \times 6.022 \times 10^{23} \, \text{nuclei/mol} \]
\[ = 3.011 \times 10^{26} \, \text{nuclei} \]

### Step 5: Calculate Total Energy Released
Since each fusion reaction releases \( 5.232 \times 10^{-13} \, \text{J} \), the total energy released from all the deuterium nuclei can be calculated as:
\[ \text{Total Energy} = \text{Number of nuclei} \times \text{Energy per reaction} \]
\[ = 3.011 \times 10^{26} \times 5.232 \times 10^{-13} \, \text{J} \]
\[ = 1.57 \times 10^{14} \, \text{J} \]

### Step 6: Approximate the Result
We can approximate this value:
\[ \approx 1.6 \times 10^{14} \, \text{J} \]

### Final Answer
The energy released in the fusion of 1 kg of deuterium nuclei is approximately:
\[ 1.6 \times 10^{14} \, \text{J} \]

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